Squares and nth powers

Using the BQD Iterator it's possible to show a connection between squares and integers raised to an arbitrarily high power.

The appropriate BQD iterator is, given:

u² + (m - 1)v² = F

then it must also be true that

(u - (m - 1)v)² + (m - 1)(u + v)² = m*F

So if you start the iterator with u = v = 1, or u = v = -1, F = m.

That means that in general there must always exist nonzero x and y, such that for an integer n equal to 0 or higher, and an integer m equal to 3 or higher:

x² + (m-1)y² = mⁿ⁺¹

I have m raised to n+1 so that n is a count of iterations, if you want m raised to the nth power you just start n at 1 instead of starting at 0, which I prefer.

Wow. That's really cool. Let's try it.

Let's let m = 5, and u = v = 1. Then, iterator is:

(u - 4v)² + 4(u + v)² = 5*F

Start is:

1² + 4*1² = 5

then it must also be true that

(-3)² + 4(2)² = 25 = 5²

Next iteration: (-11)² + 4(-1)² = 125 = 5³

And third iteration: (-7)² + 4(-12)² = 625 = 5⁴

Fourth iteration: (41)² + 4(-19)² = 3125 = 5⁵

Fifth iteration: (117)² + 4(22)² = 15625 = 5⁶

Sixth iteration: (29)² + 4(139)² = 78125 = 5⁷

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So it's easy to see how in general: x² + 4y² = 5ⁿ⁺¹

And calculating x and y is easy with the iterator. I'll note that it's important to keep up with signs. But of course x or y can be positive or negative, which means you can switch signs for either at each iteration.

That's how you can figure out each x and y that will work. But the math can also simply force in a factor of 5 into each, which means 25 divides off, pushing you backwards.

Oh yeah, and with m-1 a square you end up with a sum of squares, which is what happened with my example above, which is the first one since m must be 3 or greater. So if m - 1 is a square, then mⁿ can be written as a sum of exactly two squares for any n greater than 0. That's cool too. Wow, love these infinity results.

So it is proven that in general squares have a connection to integers raised to an arbitrarily high power.

Actually, to be more specific, it's proven that m squares where m-1 of them are the same, can always be found to sum to m raised to an arbitrarily high power.

Watching those numbers behave as predicted is a thrill unlike any other. That thrill is what you want to share.

That's pretty cool. I wonder, did anyone know that before my research? Will have to go do some web searching.


James Harris